Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 10005;
const int Maxb = 31;
struct linker {
  int x, y, id;
  linker(int x = 0, int y = 0, int id = 0) : x(x), y(y), id(id) {}
  bool operator<(const linker &b) const {
    if (y != b.y) return y > b.y;
    if (x != b.x) return x > b.x;
    return id < b.id;
  }
};
int n;
linker L[Maxn];
int xbas[Maxb];
int rval[Maxb], masktaken[Maxn];
vector<pair<int, int> > res;
int getBit(int x) {
  if (x == 0) return -1;
  for (int i = Maxb - 1; i >= 0; i--)
    if (x & 1 << i) return i;
}
void Swap(int a, int b) {
  if (a == b) return;
  res.push_back(pair<int, int>(L[a].id, L[b].id));
  res.push_back(pair<int, int>(L[b].id, L[a].id));
  res.push_back(pair<int, int>(L[a].id, L[b].id));
  swap(L[a].x, L[b].x);
  int ab = getBit(L[a].x), bb = getBit(L[b].x);
  if (ab != -1) xbas[ab] = a;
  if (bb != -1) xbas[bb] = b;
}
void writeMask(int x) {
  for (int i = 0; i < Maxb; i++)
    if (masktaken[L[x].id] & 1 << i)
      res.push_back(pair<int, int>(L[x].id, rval[i]));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &L[i].x);
    L[i].id = i;
    for (int b = Maxb - 1; b >= 0; b--)
      if (L[i].x & 1 << b)
        if (xbas[b]) {
          res.push_back(pair<int, int>(i, xbas[b]));
          L[i].x ^= L[xbas[b]].x;
        } else {
          xbas[b] = i;
          break;
        }
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &L[i].y);
    int num = L[i].y;
    L[i].y = 0;
    for (int b = Maxb - 1; b >= 0; b--)
      if (num & 1 << b)
        if (xbas[b]) {
          num ^= L[xbas[b]].x;
          L[i].y |= 1 << b;
        } else {
          printf("-1\n");
          return 0;
        }
  }
  int cur = 1;
  while (true) {
    sort(L + cur, L + n + 1);
    for (int i = cur; i <= n; i++) {
      int b = getBit(L[i].x);
      if (b != -1) xbas[b] = i;
    }
    int b = getBit(L[cur].y);
    if (b == -1) {
      for (int i = cur; i <= n; i++) {
        if (L[i].x) res.push_back(pair<int, int>(L[i].id, L[i].id));
        writeMask(i);
      }
      break;
    } else {
      Swap(cur, xbas[b]);
      writeMask(cur);
      for (int i = b - 1; i >= 0; i--)
        if (L[cur].y & 1 << i)
          res.push_back(pair<int, int>(L[cur].id, L[xbas[i]].id));
      rval[b] = L[cur].id;
      masktaken[L[cur].id] ^= 1 << b;
      for (int i = cur + 1; i <= n; i++)
        if (L[i].y & 1 << b) {
          L[i].y ^= L[cur].y;
          masktaken[L[i].id] ^= masktaken[L[cur].id];
        }
    }
    cur++;
  }
  printf("%d\n", res.size());
  for (int i = 0; i < res.size(); i++)
    printf("%d %d\n", res[i].first, res[i].second);
  return 0;
}